Question Number 36207 by MJS last updated on 30/May/18 $${x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{those}\:\mathrm{who}\:\mathrm{want}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}… \\ $$$$ \\ $$$$\left({x}−{a}−{b}\mathrm{i}\right)\left({x}−{a}+{b}\mathrm{i}\right)\left({x}−{c}−{d}\right)\left({x}−{c}+{d}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}\left({a}+{c}\right){x}^{\mathrm{3}} +\left({a}^{\mathrm{2}} +\mathrm{4}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}}…
Question Number 101742 by bemath last updated on 04/Jul/20 $$\begin{cases}{{ab}+{a}+{b}\:=\:\mathrm{5}}\\{{bc}\:+\:{b}+{c}\:=\:\mathrm{14}}\\{{ac}\:+\:{a}+{c}\:=\:\mathrm{9}}\end{cases} \\ $$$$\mathrm{find}\:{a}+{b}+{c}\:=\:\_\_\_ \\ $$ Answered by bramlex last updated on 04/Jul/20 $$\left(\mathrm{1}\right)\:{a}\:=\:\frac{\mathrm{5}−{b}}{{b}+\mathrm{1}}\:=\:\frac{\mathrm{9}−{c}}{{c}+\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\:{b}\:=\:\frac{\mathrm{14}−{c}}{{c}+\mathrm{1}}\: \\…
Question Number 167278 by mnjuly1970 last updated on 11/Mar/22 Answered by mr W last updated on 11/Mar/22 $${f}\left({x}\right)=\frac{\sqrt{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{{x}^{\mathrm{2}} +\mathrm{1}+{x}}=\frac{\mathrm{1}}{\:\sqrt{\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}}+\sqrt{\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{x}+\frac{\mathrm{1}}{{x}}}+\frac{\mathrm{1}}{\:\sqrt{{x}+\frac{\mathrm{1}}{{x}}}}} \\…
Question Number 167268 by mathlove last updated on 11/Mar/22 $${x}^{\frac{\mathrm{2}}{\:\sqrt[{\mathrm{5}}]{\mathrm{2}}}} =\mathrm{25}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$ Commented by mkam last updated on 11/Mar/22 $${x}^{\mathrm{2}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}} } =\:\mathrm{25} \\ $$$$…
Question Number 167248 by mathlove last updated on 10/Mar/22 $$\sqrt{\mathrm{3}^{{x}} }+\mathrm{1}=\mathrm{2}^{{x}} \:\:\:\:{faind}\:{x}=? \\ $$ Commented by mr W last updated on 10/Mar/22 $$\mathrm{3}^{\frac{{x}}{\mathrm{2}}} +\mathrm{1}=\mathrm{2}^{{x}} \\…
Question Number 167241 by mathlove last updated on 10/Mar/22 $$\frac{{b}}{\mathrm{11}}\:\:{is}\:{General}\:{fraction}\:\:{and}\:\mathrm{0}.\overline {\mathrm{3}{a}} \\ $$$${faind}\:\:\left({a}−\mathrm{9}{b}\right)=? \\ $$ Answered by mr W last updated on 10/Mar/22 $$\frac{{b}}{\mathrm{11}}=\mathrm{0}.\overline {\mathrm{3}{a}}…
Question Number 101693 by bemath last updated on 04/Jul/20 $$\mathrm{There}\:\mathrm{are}\:\mathrm{4}\:\mathrm{identical}\:\mathrm{mathematics} \\ $$$$\mathrm{books},\:\mathrm{2}\:\mathrm{identic}\:\mathrm{physics}\:\mathrm{books} \\ $$$$\mathrm{and}\:\mathrm{2}\:\mathrm{identical}\:\mathrm{chemistry}\:\mathrm{books} \\ $$$$.\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{compile}\: \\ $$$$\mathrm{the}\:\mathrm{eight}\:\mathrm{books}\:\mathrm{on}\:\mathrm{the}\:\mathrm{condition} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{book}\:\mathrm{are}\:\mathrm{not}\:\mathrm{mutually} \\ $$$$\mathrm{adjacent}? \\ $$ Commented…
Question Number 167231 by mathlove last updated on 10/Mar/22 Commented by cortano1 last updated on 10/Mar/22 $$\:\:\mathrm{2}^{\mathrm{x}} .\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} =\:\mathrm{10} \\ $$$$\:\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{2}^{\mathrm{x}} .\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} \right)=\mathrm{1} \\…
Question Number 36154 by SammyKT last updated on 29/May/18 $${Q}.\:\:{If}\:{x}\neq{y}\neq{z}\:\:{and}\:\:\begin{vmatrix}{{x}}&{{x}^{\mathrm{3}} }&{{x}^{\mathrm{4}} −\mathrm{1}}\\{{y}}&{{y}^{\mathrm{3}} }&{{y}^{\mathrm{4}} −\mathrm{1}}\\{{z}\:}&{{z}^{\mathrm{3}} }&{{z}^{\mathrm{4}} −\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$ \\ $$$${Prove}\:{that}\:\:{xyz}\left({xy}+{yz}+{zx}\right)=\left({x}+{y}+{z}\right) \\ $$$$ \\ $$$${please}\:{help}. \\…
Question Number 36153 by 7991 last updated on 29/May/18 $$\frac{\left({x}+{yi}−\mathrm{2}\right)^{\mathrm{2}} }{{x}−{yi}+\mathrm{1}} \\ $$ Commented by Rasheed.Sindhi last updated on 30/May/18 $$\frac{\left\{\left(\mathrm{x}−\mathrm{2}\right)+\mathrm{yi}\right\}^{\mathrm{2}} }{\left(\mathrm{x}+\mathrm{1}\right)−\mathrm{yi}}×\frac{\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{yi}}{\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{yi}} \\ $$$$\frac{\left\{\left(\mathrm{x}−\mathrm{2}\right)+\mathrm{yi}\right\}^{\mathrm{2}} \left\{\left(\mathrm{x}+\mathrm{1}\right)+\mathrm{yi}\right\}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}}…